# The Sixth Power Problem

Only using digits 0 to 9 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) once, can you form three numbers a, b and so that the sixth power of is equal to the sum of and b?

## Solution

in essence, the problem asked you to find the solution of c6 = a + b by using digits 0-9 once. Since only ten digits are used for this problem, would most likely not contain more than sеven digits. So, 4 < c < 15. Now any sixth power is congruent to 0 or to 1 modulo 9, so ≡ 0 or 1 mod 9. Sincе the sum of thе ten digits is 45, it follows that is еither 8 or 9. Therefore, has three digits. With some trial and error, I found three basic solutions.

261709 + 435 = 86
261790 + 354 = 86
261760 + 384 = 86

In each solution, the hundred’s digit of and can be interchanged without changing the value of a + b. The same goes for the ten’s digits and for the unit’s digits. Moreover, and are interchangeable. Hence, the problem has 2³ × 2 × 3 or 48 solutions.

### Posted by Edmark M. Law

My name Edmark M. Law. I work as a freelance writer, mainly writing about science and mathematics. I am an ardent hobbyist. I like to read, solve puzzles, play chess, make origami and play basketball. In addition, I dabble in magic, particularly card magic and other sleight-of-hand type magic. I live in Hong Kong. You can find me on Twitter` and Facebook. My email is edmarklaw@learnfunfacts.com

## 8 thoughts on “The Sixth Power Problem”

1. Wow, what a great solution. “Now any sixth power is congruent to 0 or to 1 modulo 9,” Now there’s something that I never knew. Terrific.

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2. Egad. If I could even understand this, it would be my sixth power….

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